Math, asked by Abhish224, 1 year ago

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If angle DBC = 70 degrees, angle BAC = 30 degrees, find angle BCD. Further, if AB = BC, find angle ECD.

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Answered by Anonymous
99
Here, ∠BCD = ∠BAC
[angles of the same segment]

⇒ ∠BDC = 30° [∵ ∠BAC = 30°]

In ∆BCD;

∠BCD + ∠DBC + ∠BDC = 180°

⇒ ∠BCD + 70° + 30° = 180° [∵ ∠DBC = 70°]

⇒ ∠BCD = 180° - 70° - 30°

⇒ ∠BCD = 80° ---- (1)

If AB = BC

then in ∆ABC --

∠ACB = ∠BAC
[angles opposite to equal sides of a triangle are equal]

⇒ ∠ACB = 30° ----- (2)

Now ∠BCD = ∠ACB + ∠ACD

⇒ 80° = 30° + ∠ACD [using (1) and (2)]

⇒ 80° - 30° = ∠ACD

⇒ 50° = ∠ACD

⇒ ∠ACD = 50°

Hence, ∠ECD = 50°

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Answered by UltimateMasTerMind
107
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Solution:-

Since, Triangle ABC & BCD are on the same base.

=> Angle BAC = Angle BDC.

=> BDC = 30°

By Angle Sum Property of Triangle BCD,

=> B + C + D = 180°

=> 70° + C + 30° = 180°

=> C = 180° - 100°

=> Angle C = 80°

Now,

Since, Triangle ABC is an ISOSCELES Triangle.

=> /_ BAC = /_ BCA

=> /_ BCA = 30°

Now,

/_ BCD = /_ BCA + /_ DCB

=> 80° = 30° + /_ DCB

=> /_ DCB = 50°

=> /_ EDC = 50°

Anonymous: Nice Ahaan!❤
Anonymous: Most welcome! ^^"
Anonymous: U r Most Welcome Ahaan!!❤
limelight1726: Nicely explained
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